What is inverse oflaplace 1/(s(s^1+1)) ?

The answer to inverse oflaplace 1/(s(s^1+1)) is \H(t)-e^{-t}

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inverse oflaplace 1/(s(s^1+1))

inverse laplace

\frac{1}{s ( s ^{1} + 1 )}

H (t) - e^{- t}

Solution steps

L^{- 1} {\frac{1}{s ( s ^{1} + 1 )}}

Take the partial fraction of

\frac{1}{s ( s ^{1} + 1 )} : \frac{1}{s} - \frac{1}{s + 1}

= L^{- 1} {\frac{1}{s} - \frac{1}{s + 1}}

Use the linearity property of Inverse Laplace Transform:

For functions

f (s), g (s)

and constants

a, b : L^{- 1} {a \cdot f (s) + b \cdot g (s)} = a \cdot L^{- 1} {f (s)} + b \cdot L^{- 1} {g (s)}

= L^{- 1} {\frac{1}{s}} - L^{- 1} {\frac{1}{s + 1}}

L^{- 1} {\frac{1}{s}} : H (t)

L^{- 1} {\frac{1}{s + 1}} : e^{- t}

= H (t) - e^{- t}

\frac{d}{d x} (2^{3})

\frac{d}{d x} (a^{2} x^{2} + b x + 49)

\int (x^{2} + x + 1) \cdot sin (x) d x

\int \frac{arctan ( 4 x )}{1 + 16 x ^{2}} d x

y = 5 x - 2 x^{2}, (- 3, - 33)

What is inverse oflaplace 1/(s(s^1+1)) ?
The answer to inverse oflaplace 1/(s(s^1+1)) is \H(t)-e^{-t}